planegeo.pdf

PROBLEMS IN PLANE AND SOLID

GEOMETRY

v.1 Plane Geometry

Viktor Prasolov

translated and edited by Dimitry Leites

Abstract. This book has no equal. The priceless treasures of elementary geometry arenowhere else exposed in so complete and at the same time transparent form. The shortsolutions take barely 1.5 2 times more space than the formulations, while still remainingcomplete, with no gaps whatsoever, although many of the problems are quite difficult. Onlythis enabled the author to squeeze about 2000 problems on plane geometry in the book ofvolume of ca 600 pages thus embracing practically all the known problems and theorems ofelementary geometry.

The book contains non-standard geometric problems of a level higher than that of theproblems usually offered at high school. The collection consists of two parts. It is based onthree Russian editions of Prasolovs books on plane geometry.

The text is considerably modified for the English edition. Many new problems are addedand detailed structuring in accordance with the methods of solution is adopted.

The book is addressed to high school students, teachers of mathematics, mathematicalclubs, and college students.

Contents

Editors preface 11From the Authors preface 12

Chapter 1. SIMILAR TRIANGLES 15Background 15Introductory problems 151. Line segments intercepted by parallel lines 152. The ratio of sides of similar triangles 173. The ratio of the areas of similar triangles 184. Auxiliary equal triangles 18* * * 195. The triangle determined by the bases of the heights 196. Similar figures 20Problems for independent study 20Solutions 21

CHAPTER 2. INSCRIBED ANGLES 33Background 33Introductory problems 331. Angles that subtend equal arcs 342. The value of an angle between two chords 353. The angle between a tangent and a chord 354. Relations between the values of an angle and the lengths of the arc and chordassociated with the angle 365. Four points on one circle 366. The inscribed angle and similar triangles 377. The bisector divides an arc in halves 388. An inscribed quadrilateral with perpendicular diagonals 399. Three circumscribed circles intersect at one point 3910. Michels point 4011. Miscellaneous problems 40Problems for independent study 41Solutions 41

CHAPTER 3. CIRCLES 57Background 57Introductory problems 581. The tangents to circles 582. The product of the lengths of a chords segments 593. Tangent circles 594. Three circles of the same radius 605. Two tangents drawn from one point 61

3

4 CONTENTS

616. Application of the theorem on triangles heights 617. Areas of curvilinear figures 628. Circles inscribed in a disc segment 629. Miscellaneous problems 6310. The radical axis 63Problems for independent study 65Solutions 65

CHAPTER 4. AREA 79Background 79Introductory problems 791. A median divides the triangleinto triangles of equal areas 792. Calculation of areas 803. The areas of the triangles into whicha quadrilateral is divided 814. The areas of the parts into whicha quadrilateral is divided 815. Miscellaneous problems 82* * * 826. Lines and curves that divide figuresinto parts of equal area 837. Formulas for the area of a quadrilateral 838. An auxiliary area 849. Regrouping areas 85Problems for independent study 86Solutions 86

CHAPTER 5. TRIANGLES 99Background 99Introductory problems 991. The inscribed and the circumscribed circles 100* * * 100* * * 1002. Right triangles 1013. The equilateral triangles 101* * * 1014. Triangles with angles of 60 and 120 1025. Integer triangles 1026. Miscellaneous problems 1037. Menelauss theorem 104* * * 1058. Cevas theorem 1069. Simsons line 10710. The pedal triangle 10811. Eulers line and the circle of nine points 10912. Brokars points 11013. Lemoines point 111

CONTENTS 5

* * * 111Problems for independent study 112Solutions 112

Chapter 6. POLYGONS 137Background 137Introductory problems 1371. The inscribed and circumscribed quadrilaterals 137* * * 138* * * 1382. Quadrilaterals 1393. Ptolemys theorem 1404. Pentagons 1415. Hexagons 1416. Regular polygons 142* * * 142* * * 1437. The inscribed and circumscribed polygons 144* * * 1448. Arbitrary convex polygons 1449. Pascals theorem 145Problems for independent study 145Solutions 146

Chapter 7. LOCI 169Background 169Introductory problems 1691. The locus is a line or a segment of a line 169* * * 1702. The locus is a circle or an arc of a circle 170* * * 1703. The inscribed angle 1714. Auxiliary equal triangles 1715. The homothety 1716. A method of loci 1717. The locus with a nonzero area 1728. Carnots theorem 1729. Fermat-Apolloniuss circle 173Problems for independent study 173Solutions 174

Chapter 8. CONSTRUCTIONS 1831. The method of loci 1832. The inscribed angle 1833. Similar triangles and a homothety 1834. Construction of triangles from various elements 1835. Construction of triangles given various points 1846. Triangles 1847. Quadrilaterals 1858. Circles 185

6 CONTENTS

9. Apollonius circle 18610. Miscellaneous problems 18611. Unusual constructions 18612. Construction with a ruler only 18613. Constructions with the help of a two-sided ruler 18714. Constructions using a right angle 188Problems for independent study 188Solutions 189

Chapter 9. GEOMETRIC INEQUALITIES 205Background 205Introductory problems 2051. A median of a triangle 2052. Algebraic problems on the triangle inequality 2063. The sum of the lengths of quadrilaterals diagonals 2064. Miscellaneous problems on the triangle inequality 207* * * 2075. The area of a triangle does not exceed a half product of two sides 2076. Inequalities of areas 2087. Area. One figure lies inside another 209* * * 2098. Broken lines inside a square 2099. The quadrilateral 21010. Polygons 210* * * 21111. Miscellaneous problems 211* * * 211Problems for independent study 212Supplement. Certain inequalities 212Solutions 213

Chapter 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE 2351. Medians 2352. Heights 2353. The bisectors 2354. The lengths of sides 2365. The radii of the circumscribed, inscribed and escribed circles 2366. Symmetric inequalities between the angles of a triangle 2367. Inequalities between the angles of a triangle 2378. Inequalities for the area of a triangle 237* * * 2389. The greater angle subtends the longer side 23810. Any segment inside a triangle is shorter than the longest side 23811. Inequalities for right triangles 23812. Inequalities for acute triangles 23913. Inequalities in triangles 239Problems for independent study 240Solutions 240

Chapter 11. PROBLEMS ON MAXIMUM AND MINIMUM 255

CONTENTS 7

Background 255Introductory problems 2551. The triangle 255* * * 2562. Extremal points of a triangle 2563. The angle 2574. The quadrilateral 2575. Polygons 2576. Miscellaneous problems 2587. The extremal properties of regular polygons 258Problems for independent study 258Solutions 259

Chapter 12. CALCULATIONS AND METRIC RELATIONS 271Introductory problems 2711. The law of sines 2712. The law of cosines 2723. The inscribed, the circumscribed and escribed circles; their radii 2724. The lengths of the sides, heights, bisectors 2735. The sines and cosines of a triangles angles 2736. The tangents and cotangents of a triangles angles 2747. Calculation of angles 274* * * 2748. The circles 275* * * 2759. Miscellaneous problems 27510. The method of coordinates 276Problems for independent study 277Solutions 277

Chapter 13. VECTORS 289Background 289Introductory problems 2891. Vectors formed by polygons (?) sides 2902. Inner product. Relations 2903. Inequalities 2914. Sums of vectors 2925. Auxiliary projections 2926. The method of averaging 2937. Pseudoinner product 293Problems for independent study 294Solutions 295

Chapter 14. THE CENTER OF MASS 307Background 3071. Main properties of the center of mass 3072. A theorem on mass regroupping 3083. The moment of inertia 3094. Miscellaneous problems 3105. The barycentric coordinates 310

8 CONTENTS

Solutions 311

Chapter 15. PARALLEL TRANSLATIONS 319Background 319Introductory problems 3191. Solving problems with the aid of parallel translations 3192. Problems on construction and loci 320* * * 320Problems for independent study 320Solutions 320

Chapter 16. CENTRAL SYMMETRY 327Background 327Introductory problems 3271. Solving problems with the help of a symmetry 3272. Properties of the symmetry 3283. Solving problems with the help of a symmetry. Constructions 328Problems for independent study 329Solutions 329

Chapter 17. THE SYMMETRY THROUGH A LINE 335Background 335Introductory problems 3351. Solving problems with the help of a symmetry 3352. Constructions 336* * * 3363. Inequalities and extremals 3364. Compositions of symmetries 3365. Properties of symmetries and axes of symmetries 3376. Chasless theorem 337Problems for independent study 338Solutions 338

Chapter 18. ROTATIONS 345Background 345Introductory problems 3451. Rotation by 90 3452. Rotation by 60 3463. Rotations through arbitrary angles 3474. Compositions of rotations 347* * * 348* * * 348Problems for independent study 348Solutions 349

Chapter 19. HOMOTHETY AND ROTATIONAL HOMOTHETY 359Background 359Introductory problems 3591. Homothetic polygons 3592. Homothetic circles 3603. Costructions and loci 360

CONTENTS 9

* * * 3614. Composition of homotheties 3615. Rotational homothety 361* * * 362* * * 3626. The center of a rotational homothety 3627. The similarity circle of three figures 363Problems for independent study 364Solutions 364

Chapter 20. THE PRINCIPLE OF AN EXTREMAL ELEMENT 375Background 3751. The least and the greatest angles 3752. The least and the greatest distances 3763. The least and the greatest areas 3764. The greatest triangle 3765. The convex hull and the base lines 3766. Miscellaneous problems 378Solutions 378

Chapter 21. DIRICHLETS PRINCIPLE 385Background 3851. The case when there are finitely many points, lines, etc. 3852. Angles and lengths 3863. Area 387Solutions 387

Chapter 22. CONVEX AND NONCONVEX POLYGONS 397Background 3971. Convex polygons 397* * * 3972. Hellys theorem 3983. Non-convex polygons 398Solutions 399

Chapter 23. DIVISIBILITY, INVARIANTS, COLORINGS 409Background 4091. Even and odd 4092. Divisibility 4103. Invariants 4104. Auxiliary colorings 4115. More auxiliary colorings 412* * * 4126. Problems on colorings 412* * * 413Solutions 413

Chapter 24. INTEGER LATTICES 4251. Polygons with vertices in the nodes of a lattice 4252. Miscellaneous problems 425Solutions 426

10 CONTENTS

Chapter 25. CUTTINGS 4311. Cuttings into parallelograms 4312. How lines cut the plane 431Solutions 432

Chapter 26. SYSTEMS OF POINTS AND SEGMENTS.EXAMPLES AND COUNTEREXAMPLES 437

1. Systems of points 4372. Systems of segments, lines and circles 4373. Examples and counterexamples 438Solutions 438

Chapter 27. INDUCTION AND COMBINATORICS 4451. Induction 4452. Combinatorics 445Solutions 445

Chapter 28. INVERSION 449Background 4491. Properties of inversions 4492. Construction of circles 4503. Constructions with the help of a compass only 4504. Let us perform an inversion 4515. Points that lie on one circle and circles passing through one point 4526. Chains of circles 454Solutions 455

Chapter 29. AFFINE TRANSFORMATIONS 4651. Affine transformations 4652. How to solve problems with the help of affine transformations 466Solutions 466

Chapter 30. PROJECTIVE TRANSFORMATIONS 4731. Projective transformations of the line 4732. Projective transformations of the plane 4743. Let us transform the given line into the infinite one 4774. Application of projective maps that preserve a circle 4785. Application of projective transformations of the line 4796. Application of projective transformations of the line in problems on construction 4797. Impossibility of construction with the help of a ruler only 480Solutions 480

Index 493

EDITORS PREFACE 11

Editors preface

The enormous number of problems and theorems of elementary geometry was consideredtoo wide to grasp in full even in the last century. Even nowadays the stream of new problemsis still wide. (The majority of these problems, however, are either well-forgotten old ones orthose recently pirated from a neighbouring country.)

Any attempt to collect an encyclopedia of all the problems seems to be doomed to failurefor many reasons.

First of all, this is an impossible task because of the huge number of the problems, anenormity too vast to grasp. Second, even if this might have been possible, the book wouldbe terribly overloaded, and therefore of no interest to anybody.

However, in the book Problems in plane geometry followed by Problems in solid geometrythis task is successfully perfomed.

In the process of writing the book the author used the books and magazines publishedin the last century as well as modern ones. The reader can judge the completeness of thebook by, for instance, the fact that American Mathematical Monthly yearly1 publishes, asnew, 12 problems already published in the Russian editions of this book.

The book turned out to be of interest to a vast audience: about 400 000 copies of thefirst edition of each of the Parts (Parts 1 and 2 Plane and Part 3 Solid) were sold;the second edition, published 5 years later, had an even larger circulation, the total over1 000 000 copies. The 3rd edition of Problems in Plane Geometry was issued in 1996 andthe latest one in 2001.

The readers interest is partly occasioned by a well-thought classification system.

The collection consists of three parts.Part 1 covers classical subjects of plane geometry. It contains nearly 1000 problems with

complete solutions and over 100 problems to be solved on ones own. Still more will be addedfor the English version of the book.

Part 2 includes more recent topics, geometric transformations and problems more suitablefor contests and for use in mathematical clubs. The problems cover cuttings, colorings, thepigeonhole (or Dirichlets) principle, induction, and so on.

Part 3 is devoted to solid geometry.

A rather detailed table of contents serves as a guide in the sea of geometric problems. Ithelps the experts to easily find what they need while the uninitiated can quickly learn whatexactly is that they are interested in in geometry. Splitting the book into small sections (5to 10 problems in each) made the book of interest to the readers of various levels.

FOR THE ENGLISH VERSION of the book about 150 new problems are already addedand several hundred more of elementary and intermideate level problems will be added tomake the number of more elementary problems sufficient to use the book in the ordinaryschool: the Russian editions are best suited for coaching for a mathematical Olympiad thanfor a regular class work: the level of difficulty increases rather fast.

Problems in each section are ordered difficulty-wise. The first problems of the sectionsare simple; they are a match for many. Here are some examples:

1Here are a few samples: v. 96, n. 5, 1989, p. 429431 (here the main idea of the solution is theright illustration precisely the picture from the back cover of the 1st Russian edition of Problems in SolidGeometry, Fig. to Problem 13.22); v. 96, n. 6, p. 527, Probl. E3192 corresponds to Problems 5.31 and18.20 of Problems in Plane Geometry with their two absolutely different solutions, the one to Problem5.31, unknown to AMM, is even more interesting.

12 CONTENTS

Plane 1.1. The bases of a trapezoid are a and b. Find the length of the segment thatthe diagonals of the trapezoid intersept on the trapezoids midline.

Plane 1.52. Let AA1 and BB1 be the altitudes of ABC. Prove that A1B1C issimilar to ABC. What is the similarity coefficient?

Plane 2.1. A line segment connects vertex A of an acute ABC with the center O ofthe circumscribed circle. The altitude AH is dropped from A. Prove that BAH = OAC.

Plane 6.1. Prove that if the center of the circle inscribed in a quadrilateral coincides withthe intersection point of the quadrilaterals diagonals, then the quadrilateral is a rhombus.

Solid 1. Arrange 6 match sticks to get 4 equilateral triangles with side length equal tothe length of a stick.

Solid 1.1. Consider the cube ABCDA1B1C1D1 with side length a. Find the angle andthe distance between the lines A1B and AC1.

Solid 6.1. Is it true that in every tetrahedron the heights meet at one point?The above problems are not difficult. The last problems in the sections are a challenge

for the specialists in geometry. It is important that the passage from simple problems tocomplicated ones is not too long; there are no boring and dull long sequences of simplesimilar problems. (In the Russian edition these sequences are, perhaps, too short, so moreproblems are added.)

The final problems of the sections are usually borrowed from scientific journals. Here aresome examples:

Plane 10.20. Prove that la+ lb+mc 3p, where la, lb are the lengths of the bisectors

of the angles A and B of the triangle ABC, mc is the length of the median of the sideAB, and p is the semiperimeter.

Plane 19.55. Let O be the center of the circle inscribed in ABC, K the Lemoinespoint, P and Q Brocards points. Prove that P and Q belong to the circle with diameterKO and that OP = OQ.

Plane 22.29. The numbers 1, . . . , n, whose sum is equal to (n2), satisfy inequalities0 < i < 2. Prove that there exists an n-gon A1 . . . An with the angles 1, . . . , n at thevertices A1, . . . , An, respectively.

Plane 24.12. Prove that for any n there exists a circle on which there lie precisely npoints with integer coordinates.

Solid 4.48. Consider several arcs of great circles on a sphere with the sum of their anglemeasures < . Prove that there exists a plane that passes through the center of the spherebut does not intersect any of these arcs.

Solid 14.22. Prove that if the centers of the escribed spheres of a tetrahedron belongto the circumscribed sphere, then the tetrahedrons faces are equal.

Solid 15.34. In space, consider 4 points not in one plane. How many various parallelip-ipeds with vertices in these points are there?

From the Authors preface

The book underwent extensive revision. The solutions to many of the problems wererewritten and about 600 new problems were added, particularly those concerning the ge-ometry of the triangle. I was greatly influenced in the process by the second edition of thebook by I. F. Sharygin Problems on Geometry. Plane geometry, Nauka, Moscow,1986 and awonderful and undeservedly forgotten book by D. Efremov New Geometry of the Triangle,Matezis, Odessa, 1902.

The present book can be used not only as a source of optional problems for studentsbut also as a self-guide for those who wish (or have no other choice but) to study geometry

FROM THE AUTHORS PREFACE 13

independently. Detailed headings are provided for the readers convenience. Problems in thetwo parts of Plane are spread over 29 Chapters, each Chapter comprising 6 to 14 sections.The classification is based on the methods used to solve geometric problems. The purpose ofthe division is basically to help the reader find his/her bearings in this large array of problems.Otherwise the huge number of problems might be somewhat depressingly overwhelming.

Advice and comments given by Academician A. V. Pogorelov, and Professors A. M. Abramov,A. Yu. Vaintrob, N. B. Vasiliev, N. P. Dolbilin, and S. Yu. Orevkov were a great help to mein preparing the first Soviet edition. I wish to express my sincere gratitude to all of them.

To save space, sections with background only contain the material directly pertinent tothe respective chapter. It is collected just to remind the reader of notations. Therefore, thebasic elements of a triangle are only defined in chapter 5, while in chapter 1 we assume thattheir definition is known. For the readers convenience, cross references in this translationare facilitated by a very detailed index.

Chapter 1. SIMILAR TRIANGLES

Background

1) Triangle ABC is said to be similar to triangle A1B1C1 (we writeABC A1B1C1)if and only if one of the following equivalent conditions is satisfied:

a) AB : BC : CA = A1B1 : B1C1 : C1A1;b) AB : BC = A1B1 : B1C1 and ABC = A1B1C1;c) ABC = A1B1C1 and BAC = B1A1C1.2) Triangles AB1C1 and AB2C2 cut off from an angle with vertex A by parallel lines are

similar and AB1 : AB2 = AC1 : AC2 (here points B1 and B2 lie on one leg of the angle andC1 and C2 on the other leg).

3) A midline of a triangle is the line connecting the midpoints of two of the trianglessides. The midline is parallel to the third side and its length is equal to a half length of thethird side.

The midline of a trapezoid is the line connecting the midpoints of the trapezoids sides.This line is parallel to the bases of the trapezoid and its length is equal to the halfsum oftheir lengths.

4) The ratio of the areas of similar triangles is equal to the square of the similaritycoefficient, i.e., to the squared ratio of the lengths of respective sides. This follows, forexample, from the formula SABC =

12AB AC sinA.

5) Polygons A1A2 . . . An and B1B2 . . . Bn are called similar if A1A2 : A2A3 : : AnA1 =B1B2 : B2B3 : : BnB1 and the angles at the vertices A1, . . . , An are equal to the anglesat the vertices B1, . . . , Bn, respectively.

The ratio of the respective diagonals of similar polygons is equal to the similarity coeffi-cient. For the circumscribed similar polygons, the ratio of the radii of the inscribed circlesis also equal to the similarity coefficient.

Introductory problems

1. Consider heights AA1 and BB1 in acute triangle ABC. Prove that A1C BC =B1C AC.

2. Consider height CH in right triangle ABC with right angle C. Prove that AC2 =AB AH and CH2 = AH BH.

3. Prove that the medians of a triangle meet at one point and this point divides eachmedian in the ratio of 2 : 1 counting from the vertex.

4. On side BC of ABC point A1 is taken so that BA1 : A1C = 2 : 1. What is theratio in which median CC1 divides segment AA1?

5. Square PQRS is inscribed into ABC so that vertices P and Q lie on sides AB andAC and vertices R and S lie on BC. Express the length of the squares side through a andha.

1. Line segments intercepted by parallel lines

1.1. Let the lengths of bases AD and BC of trapezoid ABCD be a and b (a > b).

15

16 CHAPTER 1. SIMILAR TRIANGLES

a) Find the length of the segment that the diagonals intercept on the midline.b) Find the length of segment MN whose endpoints divide AB and CD in the ratio of

AM :MB = DN : NC = p : q.1.2. Prove that the midpoints of the sides of an arbitrary quadrilateral are vertices of

a parallelogram. For what quadrilaterals this parallelogram is a rectangle, a rhombus, asquare?

1.3. Points A1 and B1 divide sides BC and AC ofABC in the ratios BA1 : A1C = 1 : pand AB1 : B1C = 1 : q, respectively. In what ratio is AA1 divided by BB1?

1.4. Straight lines AA1 and BB1 pass through point P of median CC1 in ABC (A1and B1 lie on sides BC and CA, respectively). Prove that A1B1 AB.

1.5. The straight line which connects the intersection point P of the diagonals in quadri-lateral ABCD with the intersection point Q of the lines AB and CD bisects side AD. Provethat it also bisects BC.

1.6. A point P is taken on side AD of parallelogram ABCD so that AP : AD = 1 : n;let Q be the intersection point of AC and BP . Prove that AQ : AC = 1 : (n+ 1).

1.7. The vertices of parallelogram A1B1C1D1 lie on the sides of parallelogram ABCD(point A1 lies on AB, B1 on BC, etc.). Prove that the centers of the two parallelogramscoincide.

1.8. Point K lies on diagonal BD of parallelogram ABCD. Straight line AK intersectslines BC and CD at points L and M , respectively. Prove that AK2 = LK KM .

1.9. One of the diagonals of a quadrilateral inscribed in a circle is a diameter of thecircle. Prove that (the lengths of) the projections of the opposite sides of the quadrilateralon the other diagonal are equal.

1.10. Point E on base AD of trapezoid ABCD is such that AE = BC. Segments CAand CE intersect diagonal BD at O and P , respectively. Prove that if BO = PD, thenAD2 = BC2 + AD BC.

1.11. On a circle centered at O, points A and B single out an arc of 60. Point Mbelongs to this arc. Prove that the straight line passing through the midpoints of MA andOB is perpendicular to that passing through the midpoints of MB and OA.

1.12. a) Points A, B, and C lie on one straight line; points A1, B1, and C1 lie on anotherstraight line. Prove that if AB1 BA1 and AC1 CA1, then BC1 CB1.

b) Points A, B, and C lie on one straight line and A1, B1, and C1 are such that AB1 BA1, AC1 CA1, and BC1 CB1. Prove that A1, B1 and C1 lie on one line.

1.13. In ABC bisectors AA1 and BB1 are drawn. Prove that the distance from anypoint M of A1B1 to line AB is equal to the sum of distances from M to AC and BC.

1.14. Let M and N be the midpoints of sides AD and BC in rectangle ABCD. PointP lies on the extension of DC beyond D; point Q is the intersection point of PM and AC.Prove that QNM = MNP .

1.15. Points K and L are taken on the extensions of the bases AD and BC of trapezoidABCD beyond A and C, respectively. Line segment KL intersects sides AB and CD at Mand N , respectively; KL intersects diagonals AC and BD at O and P , respectively. Provethat if KM = NL, then KO = PL.

1.16. Points P , Q, R, and S on sides AB, BC, CD and DA, respectively, of convexquadrilateralABCD are such thatBP : AB = CR : CD = andAS : AD = BQ : BC = .Prove that PR and QS are divided by their intersection point in the ratios : (1 ) and : (1 ), respectively.

2. THE RATIO OF SIDES OF SIMILAR TRIANGLES 17

2. The ratio of sides of similar triangles

1.17. a) In ABC bisector BD of the external or internal angle B is drawn. Provethat AD : DC = AB : BC.

b) Prove that the center O of the circle inscribed in ABC divides the bisector AA1 inthe ratio of AO : OA1 = (b+ c) : a, where a, b and c are the lengths of the triangles sides.

1.18. The lengths of two sides of a triangle are equal to a while the length of the thirdside is equal to b. Calculate the radius of the circumscribed circle.

1.19. A straight line passing through vertex A of square ABCD intersects side CD atE and line BC at F . Prove that 1

AE2+ 1

AF 2= 1

AB2.

1.20. Given points B2 and C2 on heights BB1 and CC1 of ABC such that AB2C =AC2B = 90

, prove that AB2 = AC2.1.21. A circle is inscribed in trapezoid ABCD (BC AD). The circle is tangent to sides

AB and CD at K and L, respectively, and to bases AD and BC at M and N , respectively.a) Let Q be the intersection point of BM and AN . Prove that KQ AD.b) Prove that AK KB = CL LD.1.22. Perpendiculars AM and AN are dropped to sides BC and CD of parallelogram

ABCD (or to their extensions). Prove that MAN ABC.1.23. Straight line l intersects sides AB and AD of parallelogram ABCD at E and F ,

respectively. Let G be the intersection point of l with diagonal AC. Prove that ABAE

+ ADAF

=ACAG

.1.24. Let AC be the longer of the diagonals in parallelogram ABCD. Perpendiculars

CE and CF are dropped from C to the extensions of sides AB and AD, respectively. Provethat AB AE + AD AF = AC2.

1.25. Angles and of ABC are related as 3+2 = 180. Prove that a2 + bc = c2.1.26. The endpoints of segments AB and CD are gliding along the sides of a given angle,

so that straight lines AB and CD are moving parallelly (i.e., each line moves parallelly toitself) and segments AB and CD intersect at a point, M . Prove that the value of AM BM

CM DM isa constant.

1.27. Through an arbitrary point P on side AC of ABC straight lines are drawnparallelly to the triangles medians AK and CL. The lines intersect BC and AB at E andF , respectively. Prove that AK and CL divide EF into three equal parts.

1.28. Point P lies on the bisector of an angle with vertex C. A line passing through Pintercepts segments of lengths a and b on the angles legs. Prove that the value of 1

a+ 1

bdoes

not depend on the choice of the line.1.29. A semicircle is constructed outwards on side BC of an equilateral triangle ABC

as on the diameter. Given points K and L that divide the semicircle into three equal arcs,prove that lines AK and AL divide BC into three equal parts.

1.30. Point O is the center of the circle inscribed in ABC. On sides AC and BCpoints M and K, respectively, are selected so that BK AB = BO2 and AM AB = AO2.Prove that M , O and K lie on one straight line.

1.31. Equally oriented similar triangles AMN , NBM and MNC are constructed onsegment MN (Fig. 1).

Prove that ABC is similar to all these triangles and the center of its curcumscribedcircle is equidistant from M and N .

1.32. Line segment BE divides ABC into two similar triangles, their similarity ratiobeing equal to

3.

Find the angles of ABC.


Figure 1 (1.31)

3. The ratio of the areas of similar triangles

1.33. A point E is taken on side AC of ABC. Through E pass straight lines DEand EF parallel to sides BC and AB, respectively; D and E are points on AB and BC,respectively. Prove that SBDEF = 2

SADE SEFG.

1.34. PointsM and N are taken on sides AB and CD, respectively, of trapezoid ABCDso that segment MN is parallel to the bases and divides the area of the trapezoid in halves.Find the length of MN if BC = a and AD = b.

1.35. Let Q be a point inside ABC. Three straight lines are pass through Q par-allelly to the sides of the triangle. The lines divide the triangle into six parts, three ofwhich are triangles of areas S1, S2 and S3. Prove that the area of ABC is equal to(

S1 +S2 +

S3)2.

1.36. Prove that the area of a triangle whose sides are equal to the medians of a triangleof area S is equal to 3

4S.

1.37. a) Prove that the area of the quadrilateral formed by the midpoints of the sides ofconvex quadrilateral ABCD is half that of ABCD.

b) Prove that if the diagonals of a convex quadrilateral are equal, then its area is theproduct of the lengths of the segments which connect the midpoints of its opposite sides.

1.38. Point O lying inside a convex quadrilateral of area S is reflected symmetricallythrough the midpoints of its sides. Find the area of the quadrilateral with its vertices in theimages of O under the reflections.

4. Auxiliary equal triangles

1.39. In right triangle ABC with right angle C, points D and E divide leg BC of intothree equal parts. Prove that if BC = 3AC, then AEC + ADC + ABC = 90.

1.40. Let K be the midpoint of side AB of square ABCD and let point L divide diagonalAC in the ratio of AL : LC = 3 : 1. Prove that KLD is a right angle.

1.41. In square ABCD straight lines l1 and l2 pass through vertex A. The lines intersectthe squares sides. Perpendiculars BB1, BB2, DD1, and DD2 are dropped to these lines.Prove that segments B1B2 and D1D2 are equal and perpendicular to each other.

1.42. Consider an isosceles right triangle ABC with CD = CE and points D and E onsides CA and CB, respectively. Extensions of perpendiculars dropped from D and C to AEintersect the hypotenuse AB at K and L. Prove that KL = LB.

1.43. Consider an inscribed quadrilateral ABCD. The lengths of sides AB, BC, CD,and DA are a, b, c, and d, respectively. Rectangles are constructed outwards on the sides of

5. THE TRIANGLE DETERMINED BY THE BASES OF THE HEIGHTS 19

the quadrilateral; the sizes of the rectangles are a c, b d, c a and d b, respectively.Prove that the centers of the rectangles are vertices of a rectangle.

1.44. Hexagon ABCDEF is inscribed in a circle of radius R centered at O; let AB =CD = EF = R. Prove that the intersection points, other than O, of the pairs of circlescircumscribed about BOC, DOE and FOA are the vertices of an equilateral trianglewith side R.

* * *

1.45. Equilateral triangles BCK and DCL are constructed outwards on sides BC andCD of parallelogram ABCD. Prove that AKL is an equilateral triangle.

1.46. Squares are constructed outwards on the sides of a parallelogram. Prove that theircenters form a square.

1.47. Isosceles triangles with angles 2, 2 and 2 at vertices A, B and C are con-structed outwards on the sides of triangle ABC; let ++ = 180. Prove that the anglesof ABC are equal to , and .

1.48. On the sides of ABC as on bases, isosceles similar triangles AB1C and AC1Bare constructed outwards and an isosceles triangle BA1C is constructed inwards. Prove thatAB1A1C1 is a parallelogram.

1.49. a) On sides AB and AC of ABC equilateral triangles ABC1 and AB1C areconstructed outwards; let C1 = B1 = 90

, ABC1 = ACB1 = ; let M be themidpoint of BC. Prove that MB1 =MC1 and B1MC1 = 2.

b) Equilateral triangles are constructed outwards on the sides of ABC. Prove that thecenters of the triangles constructed form an equilateral triangle whose center coincides withthe intersection point of the medians of ABC.

1.50. Isosceles triangles AC1B and AB1C with an angle at the vertex are constructedoutwards on the unequal sides AB and AC of a scalene triangle ABC.

a) Let M be a point on median AA1 (or on its extension), let M be equidistant from B1and C1. Prove that B1MC1 = .

b) Let O be a point of the midperpendicular to segment BC, let O be equidistant fromB1 and C1. Prove that B1OC = 180

.1.51. Similar rhombuses are constructed outwards on the sides of a convex rectangle

ABCD, so that their acute angles (equal to ) are adjacent to vertices A and C. Provethat the segments which connect the centers of opposite rhombuses are equal and the anglebetween them is equal to .

5. The triangle determined by the bases of the heights

1.52. Let AA1 and BB1 be heights of ABC. Prove that A1B1C ABC. Whatis the similarity coefficient?

1.53. Height CH is dropped from vertex C of acute triangle ABC and perpendicularsHM and HN are dropped to sides BC and AC, respectively. Prove thatMNC ABC.

1.54. In ABC heights BB1 and CC1 are drawn.a) Prove that the tangent at A to the circumscribed circle is parallel to B1C1.b) Prove that B1C1 OA, where O is the center of the circumscribed circle.1.55. Points A1, B1 and C1 are taken on the sides of an acute triangle ABC so that

segments AA1, BB1 and CC1 meet at H. Prove that AH A1H = BH B1H = CH C1Hif and only if H is the intersection point of the heights of ABC.

1.56. a) Prove that heights AA1, BB1 and CC1 of acute triangle ABC bisect the anglesof A1B1C1.


b) Points C1, A1 andB1 are taken on sidesAB, BC and CA, respectively, of acute triangleABC. Prove that if B1A1C = BA1C1, A1B1C = AB1C1 and A1C1B = AC1B1,then points A1, B1 and C1 are the bases of the heights of ABC.

1.57. Heights AA1, BB1 and CC1 are drawn in acute triangle ABC. Prove that thepoint symmetric to A1 through AC lies on B1C1.

1.58. In acute triangle ABC, heights AA1, BB1 and CC1 are drawn. Prove that ifA1B1 AB and B1C1 BC, then A1C1 AC.

1.59. Let p be the semiperimeter of acute triangle ABC and q the semiperimeter of thetriangle formed by the bases of the heights of ABC. Prove that p : q = R : r, where Rand r are the radii of the circumscribed and the inscribed circles, respectively, of ABC.

6. Similar figures

1.60. A circle of radius r is inscribed in a triangle. The straight lines tangent to thecircle and parallel to the sides of the triangle are drawn; the lines cut three small trianglesoff the triangle. Let r1, r2 and r3 be the radii of the circles inscribed in the small triangles.Prove that r1 + r2 + r3 = r.

1.61. Given ABC, draw two straight lines x and y such that the sum of lengths ofthe segments MXM and MYM drawn parallel to x and y from a point M on AC to theirintersections with sides AB and BC is equal to 1 for any M .

1.62. In an isosceles triangle ABC perpendicular HE is dropped from the midpoint ofbase BC to side AC. Let O be the midpoint of HE. Prove that lines AO and BE areperpendicular to each other.

1.63. Prove that projections of the base of a triangles height to the sides between whichit lies and on the other two heights lie on the same straight line.

1.64. Point B lies on segment AC; semicircles S1, S2, and S3 are constructed on one sideof AC, as on diameter. Let D be a point on S3 such that BD AC. A common tangentline to S1 and S2 touches these semicircles at F and E, respectively.

a) Prove that EF is parallel to the tangent to S3 passing through D.b) Prove that BFDE is a rectangle.1.65. Perpendiculars MQ and MP are dropped from an arbitrary point M of the circle

circumscribed about rectangle ABCD to the rectangles two opposite sides; the perpendic-ulars MR and MT are dropped to the extensions of the other two sides. Prove that linesPR QT and the intersection point of PR and QT belongs to a diagonal of ABCD.

1.66. Two circles enclose non-intersecting areas. Common tangent lines to the twocircles, one external and one internal, are drawn. Consider two straight lines each of whichpasses through the tangent points on one of the circles. Prove that the intersection point ofthe lines lies on the straight line that connects the centers of the circles.

Problems for independent study

1.67. The (length of the) base of an isosceles triangle is a quarter of its perimeter. Froman arbitrary point on the base straight lines are drawn parallel to the sides of the triangle.How many times is the perimeter of the triangle greater than that of the parallelogram?

1.68. The diagonals of a trapezoid are mutually perpendicular. The intersection pointdivides the diagonals into segments. Prove that the product of the lengths of the trapezoidsbases is equal to the sum of the products of the lengths of the segments of one diagonal andthose of another diagonal.

1.69. A straight line is drawn through the center of a unit square. Calculate the sum ofthe squared distances between the four vertices of the square and the line.

SOLUTIONS 21

1.70. Points A1, B1 and C1 are symmetric to the center of the circumscribed circle ofABC through the triangles sides. Prove that ABC = A1B1C1.

1.71. Prove that if BAC = 2ABC, then BC2 = (AC + AB)AC.1.72. Consider points A, B, C and D on a line l. Through A, B and through C, D

parallel straight lines are drawn. Prove that the diagonals of the parallelograms thus formed(or their extensions) intersect l at two points that do not depend on parallel lines but dependon points A, B, C, D only.

1.73. In ABC bisector AD and midline A1C1 are drawn. They intersect at K. Provethat 2A1K = |b c|.

1.74. Points M and N are taken on sides AD and CD of parallelogram ABCD suchthat MN AC. Prove that SABM = SCBN .

1.75. On diagonal AC of parallelogram ABCD points P and Q are taken so thatAP = CQ. Let M be such that PM AD and QM AB. Prove that M lies on diagonalBD.

1.76. Consider a trapezoid with bases AD and BC. Extensions of the sides of ABCDmeet at point O. Segment EF is parallel to the bases and passes through the intersectionpoint of the diagonals. The endpoints of EF lie on AB and CD. Prove that AE : CF =AO : CO.

1.77. Three straight lines parallel to the sides of the given triangle cut three triangles offit leaving an equilateral hexagon. Find the length of the side of the hexagon if the lengthsof the triangles sides are a, b and c.

1.78. Three straight lines parallel to the sides of a triangle meet at one point, the sidesof the triangle cutting off the line segments of length x each. Find x if the lengths of thetriangles sides are a, b and c.

1.79. Point P lies inside ABC and ABP = ACP . On straight lines AB and AC,points C1 and B1 are taken so that BC1 : CB1 = CP : BP . Prove that one of the diagonalsof the parallelogram whose two sides lie on lines BP and CP and two other sides (or theirextensions) pass through B1 and C1 is parallel to BC.

Solutions

1.1. a) Let P and Q be the midpoints of AB and CD; let K and L be the intersectionpoints of PQ with the diagonals AC and BD, respectively. Then PL = a

2and PK = 1

2b

and so KL = PL PK = 12(a b).

b) Take point F on AD such that BF CD. Let E be the intersection point of MNwith BF . Then

MN =ME + EN =

q AFp+ q

+ b =q(a b) + (p+ q)b

p+ q=qa+ pb

p+ q.

1.2. Consider quadrilateral ABCD. Let K, L, M and N be the midpoints of sides AB,BC, CD and DA, respectively. Then KL =MN = 1

2AC and KL MN , that is KLMN is

a parallelogram. It becomes clear now that KLMN is a rectangle if the diagonals AC andBD are perpendicular, a rhombus if AC = BD, and a square if AC and BD are of equallength and perpendicular to each other.

1.3. Denote the intersection point of AA1 with BB1 by O. In B1BC draw segmentA1A2 so that A1A2 BB1. Then B1CB1A2 = 1 + p and so AO : OA1 = AB1 : B1A2 = B1C :qB1A2 = (1 + p) : q.


1.4. Let A2 be the midpoint of A1B. Then CA1 : A1A2 = CP : PC1 and A1A2 : A1B =1 : 2. So CA1 : A1B = CP : 2PC1. Similarly, CB1 : B1A = CP : 2PC1 = CA1 : A1B.

1.5. Point P lies on the median QM of AQD (or on its extension). It is easy toverify that the solution of Problem 1.4 remains correct also for the case when P lies on theextension of the median. Consequently, BC AD.

1.6. We have AQ : QC = AP : BC = 1 : n because AQP CQB. So AC =AQ+QC = (n+ 1)AQ.

1.7. The center of A1B1C1D1 being the midpoint of B1D1 belongs to the line segmentwhich connects the midpoints of AB and CD. Similarly, it belongs to the segment whichconnects the midpoints of BC and AD. The intersection point of the segments is the centerof ABCD.

1.8. Clearly, AK : KM = BK : KD = LK : AK, that is AK2 = LK KM .1.9. Let AC be the diameter of the circle circumscribed about ABCD. Drop perpen-

diculars AA1 and CC1 to BD (Fig. 2).

Figure 2 (Sol. 1.9)

We must prove that BA1 = DC1. Drop perpendicular OP from the center O of thecircumscribed circle to BD. Clearly, P is the midpoint of BD. Lines AA1, OP and CC1 areparallel to each other and AO = OC. So A1P = PC1 and, since P is the midpoint of BD,it follows that BA1 = DC1.

1.10. We see that BO : OD = DP : PB = k, because BO =PD. Let BC = 1. ThenAD = k and ED = 1

k. So k = AD = AE +ED = 1+ 1

k, that is k2 = 1+ k. Finally, observe

that k2 = AD2 and 1 + k = BC2 +BC AD.1.11. Let C, D, E and F be the midpoints of sides AO, OB, BM andMA, respectively,

of quadrilateral AOMB. Since AB = MO = R, where R is the radius of the given circle,CDEF is a rhombus by Problem 1.2. Hence, CE DF .

1.12. a) If the lines containing the given points are parallel, then the assertion of theproblem is obviously true. We assume that the lines meet at O. Then OA : OB = OB1 : OA1and OC : OA = OA1 : OC1. Hence, OC : OB = OB1 : OC1 and so BC1 CB1 (the ratiosof the segment should be assumed to be oriented).

b) Let AB1 and CA1 meet at D, let CB1 and AC1 meet at E. Then CA1 : A1D = CB :BA = EC1 : C1A. Since CB1D EB1A, points A1, B1 and C1 lie on the same line.

1.13. A point that lies on the bisector of an angle is equidistant from the angles legs.Let a be the distance from point A1 to lines AC and AB, let b be the distance from point B1to lines AB and BC. Further, let A1M : B1M = p : q, where p+ q = 1. Then the distances

SOLUTIONS 23

from point M to lines AC and BC are equal to qa and pb, respectively. On the other hand,by Problem 1.1 b) the distance from point M to line AB is equal to qa+ pb.

1.14. Let the line that passes through the center O of the given rectangle parallel to BCintersect line segment QN at point K (Fig. 3).

Figure 3 (Sol. 1.14)

Since MO PC, it follows that QM : MP = QO : OC and, since KO BC, it followsthat QO : OC = QK : KN . Therefore, QM : MP = QK : KN , i.e., KM NP . Hence,MNP = KMO = QNM .

1.15. Let us draw through point M line EF so that EF CD (points E and F lie onlines BC and AD). Then PL : PK = BL : KD and OK : OL = KA : CL = KA : KF =BL : EL. Since KD = EL, we have PL : PK = OK : OL and, therefore, PL = OK.

1.16. Consider parallelogram ABCD1. We may assume that points D and D1 do notcoincide (otherwise the statement of the problem is obvious). On sides AD1 and CD1 takepoints S1 and R1, respectively, so that SS1 DD1 and RR1 DD1. Let segments PR1 andQS1 meet at N ; let N1 and N2 be the intersection points of the line that passes through Nparallel to DD1 with segments PR and QS, respectively.

ThenN1N =

RR1 =

DD1 and

N2N =

SS1 =

DD1. Hence, segments PR and

QS meet at N1 = N2. Clearly, PN1 : PR = PN : PR1 = and QN2 : QS = .

Remark. If = , there is a simpler solution. Since BP : BA = BQ : BC = , itfollows that PQ AC and PQ : AC = . Similarly, RS AC and RS : AC = 1 .Therefore, segments PR and QS are divided by their intersection point in the ratio of : (1 ).

1.17. a) From vertices A and C drop perpendiculars AK and CL to line BD. SinceCBL = ABK and CDL = KDA, we see that BLC BKA and CLD AKD. Therefore, AD : DC = AK : CL = AB : BC.

b) Taking into account that BA1 : A1C = BA : AC and BA1 + A1C = BC we getBA1 =

acb+c

. Since BO is the bisector of triangle ABA1, it follows that AO : OA1 = AB :BA1 = (b+ c) : a.

1.18. Let O be the center of the circumscribed circle of isosceles triangle ABC, let B1be the midpoint of base AC and A1 the midpoint of the lateral side BC. Since BOA1 BCB1, it follows that BO : BA1 = BC : BB1 and, therefore, R = BO = a24a2b2 .


1.19. If EAD = , then AE = ADcos

= ABcos

and AF = ABsin

. Therefore,

1

AE2+

1

AF 2=

cos2 + sin2

AB2=

1

AB2.

1.20. It is easy to verify that AB22 = AB1 AC = AC1 AB = AC22 .1.21. a) Since BQ : QM = BN : AM = BK : AK, we have: KQ AM .b) Let O be the center of the inscribed circle. Since CBA+ BAD = 180, it follows

that ABO + BAO = 90. Therefore, AKO OKB, i.e., AK : KO = OK : KB.Consequently, AK KB = KO2 = R2, where R is the radius of the inscribed circle. Similarly,CL LD = R2.

1.22. If angle ABC is obtuse (resp. acute), then angle MAN is also obtuse (resp.acute). Moreover, the legs of these angles are mutually perpendicular. Therefore, ABC =MAN . Right triangles ABM and ADN have equal angles ABM = ADN , therefore,AM : AN = AB : AD = AB : CB, i.e., ABC MAN .

1.23. On diagonal AC, take points D and B such that BB l and DD l. ThenAB : AE = AB : AG and AD : AF = AD : AG. Since the sides of triangles ABB

and CDD are pairwise parallel and AB = CD, these triangles are equal and AB = CD.Therefore,

AB

AE+AD

AF=AB

AG+AD

AG=CD + AD

AG=AC

AG.

1.24. Let us drop from vertex B perpendicular BG to AC (Fig. 4).


Since triangles ABG and ACE are similar, AC AG = AE AB. Lines AF and CB areparallel, consequently, GCB = CAF . We also infer that right triangles CBG and ACFare similar and, therefore, AC CG = AF BC. Summing the equalities obtained we get

AC (AG+ CG) = AE AB + AF BC.Since AG+ CG = AC, we get the equality desired.

1.25. Since + = 90 12, it follows that = 180 = 90 + 1

2. Therefore,

it is possible to find point D on side AB so that ACD = 90 12, i.e., AC = AD. Then

ABC CBD and, therefore, BC : BD = AB : CB, i.e., a2 = c(c b).1.26. As segments AB and CD move, triangle AMC is being replaced by another triangle

similar to the initial one. Therefore, the quantity AMCM

remains a constant. Analogously, BMDM

remains a constant.1.27. Let medians meet at O; denote the intersection points of median AK with lines

FP and FE by Q and M , respectively; denote the intersection points of median CL withlines EP and FE by R and N , respectively (Fig. 5).

Clearly, FM : FE = FQ : FP = LO : LC = 1 : 3, i.e., FM = 13FE. Similarly,

EN = 13FE.

SOLUTIONS 25


1.28. Let A and B be the intersection points of the given line with the angles legs.On segments AC and BC, take points K and L, respectively, so that PK BC andPL AC. Since AKP PLB, it follows that AK : KP = PL : LB and, therefore,(a p)(b p) = p2, where p = PK = PL. Hence, 1

a+ 1

b= 1

p.

1.29. Denote the midpoint of side BC by O and the intersection points of AK and ALwith side BC by P and Q, respectively. We may assume that BP < BQ. Triangle LCOis an equilateral one and LC AB. Therefore, ABQ LCQ, i.e., BQ : QC = AB :LC = 2 : 1. Hence, BC = BQ+QC = 3QC. Similarly, BC = 3BP .

1.30. Since BK : BO = BO : AB and KBO = ABO, it follows that KOB OAB. Hence, KOB = OAB. Similarly, AOM = ABO. Therefore,

KOM = KOB + BOA+ AOM = OAB + BOA+ ABO = 180,

i.e., points K, O and M lie on one line.1.31. Since AMN = MNC and BMN = MNA, we see that AMB = ANC.

Moreover, AM : AN = NB : NM = BM : CN . Hence, AMB ANC and, therefore,MAB = NAC. Consequently, BAC = MAN . For the other angles the proof issimilar.

Let points B1 and C1 be symmetric to B and C, respectively, through the midperpen-dicular to segment MN . Since AM : NB = MN : BM = MC : NC, it follows thatMA MC1 = AM NC = NB MC = MB1 MC. Therefore, point A lies on the circlecircumscribed about trapezoid BB1CC1.

1.32. Since AEB+BEC = 180, angles AEB and BEC cannot be different anglesof similar triangles ABE and BEC, i.e., the angles are equal and BE is a perpendicular.

Two cases are possible: either ABE = CBE or ABE = BCE. The first caseshould be discarded because in this case ABE = CBE.

In the second case we have ABC = ABE + CBE = ABE + BAE = 90. Inright triangle ABC the ratio of the legs lengths is equal to 1 :

3; hence, the angles of

triangle ABC are equal to 90, 60, 30.

1.33. We have SBDEF2SADE

= SBDESADE

= DBAD

= EFAD

=

SEFCSADE

. Hence,

SBDEF = 2SADE SEFC .

1.34. Let MN = x; let E be the intersection point of lines AB and CD. TrianglesEBC, EMN and EAD are similar, hence, SEBC : SEMN : SEAD = a

2 : x2 : b2. SinceSEMN SEBC = SMBCN = SMADN = SEAD SEMN , it follows that x2 a2 = b2 x2, i.e.,x2 = 1

2(a2 + b2).


1.35. Through point Q inside triangle ABC draw lines DE, FG and HI parallel to BC,CA and AB, respectively, so that points F and H would lie on side BC, points E and I onside AC, points D and G on side AB (Fig. 6).


Set S = SABC , S1 = SGDQ, S2 = SIEQ, S3 = SHFQ. ThenS1S

+

S2S

+

S3S

=GQ

AC+

IE

AC+FQ

AC=AI + IE + EC

AC= 1,

i.e., S = (S1 +

S2 +

S3)

2.1.36. Let M be the intersection point of the medians of triangle ABC; let point A1 be

symmetric to M through the midpoint of segment BC. The ratio of the lengths of sidesof triangle CMA1 to the lengths of the corresponding medians of triangle ABC is to 2 : 3.Therefore, the area to be found is equal to 9

4SCMA1 . Clearly, SCMA1 =

13S (cf. the solution

of Problem 4.1).1.37. Let E, F , G and H be the midpoints of sides AB, BC, CD and DA, respectively.a) Clearly, SAEH + SCFG =

14SABD +

14SCBD =

14SABCD. Analogously, SBEF + SDGH =

14SABCD; hence, SEFGH = SABCD 14SABCD 14SABCD = 12SABCD.b) Since AC = BD, it follows that EFGH is a rhombus (Problem 1.2). By heading a)

we have SABCD = 2SEFGH = EG FH.1.38. Let E, F , G and H be the midpoints of sides of quadrilateral ABCD; let points

E1, F1, G1 and H1 be symmetric to point O through these points, respectively. Since EFis the midline of triangle E1OF1, we see that SE1OF1 = 4SEOF . Similarly, SF1OG1 = 4SFOG,SG1OH1 = 4SGOH , SH1OE1 = 4SHOE. Hence, SE1F1G1H1 = 4SEFGH . By Problem 1.37 a)SABCD = 2SEFGH . Hence, SE1F1G1H1 = 2SABCD = 2S.

1.39. First solution. Let us consider square BCMN and divide its sideMN by pointsP and Q into three equal parts (Fig. 7).

Then ABC = PDQ and ACD = PMA. Hence, triangle PAD is an isoscelesright triangle and ABC + ADC = PDQ+ ADC = 45.

Second solution. Since DE = 1, EA =2, EB = 2, AD =

5 and BA =

10,

it follows that DE : AE = EA : EB = AD : BA and DEA AEB. Therefore,ABC = EAD. Moreover, AEC = CAE = 45. Hence,

ABC + ADC + AEC = (EAD + CAE) + ADC

= CAD + ADC = 90.

SOLUTIONS 27


1.40. From point L drop perpendiculars LM and LN on AB and AD, respectively.Then KM = MB = ND and KL = LB = DL and, therefore, right triangles KML andDNL are equal. Hence, DLK = NLM = 90.

1.41. Since D1A = B1B,AD2 = BB2 and D1AD2 = B1BB2, it follows thatD1AD2 = B1BB2. Sides AD1 and BB1 (and also AD2 and BB2) of these trianglesare perpendicular and, therefore, B1B2 D1D2.

1.42. On the extension of segment AC beyond point C take pointM so that CM = CE(Fig. 8).


Then under the rotation with center C through an angle of 90 triangle ACE turns intotriangle BCM . Therefore, line MB is perpendicular to line AE; hence, it is parallel to lineCL. Since MC = CE = DC and lines DK, CL and MB are parallel, KL = LB.

1.43. Let rectangles ABC1D1 and A2BCD2 be constructed on sides AB and BC; letP , Q, R and S be the centers of rectangles constructed on sides AB, BC, CD and DA,respectively. Since ABC + ADC = 180, it follows that ADC = A2BC1 and, there-fore, RDS = PBQ and RS = PQ. Similarly, QR = PS. Therefore, PQRS is aparallelogram such that one of triangles RDS and PBQ is constructed on its sides outwardsand on the other side inwards; a similar statement holds for triangles QCR and SAP aswell. Therefore, PQR + RSP = BQC + DSA = 180 because PQB = RSD andRQC = PSA. It follows that PQRS is a rectangle.

1.44. Let K, L and M be the intersection points of the circumscribed circles of trianglesFOA and BOC, BOC and DOE, DOE and FOA, respectively; 2, 2 and 2 the anglesat the vertices of isosceles triangles BOC, DOE and FOA, respectively (Fig. 9).



Point K lies on arc OB of the circumscribed circle of the isosceles triangle BOC and,therefore, OKB = 90 + . Similarly, OKA = 90 + . Since + + = 90, it followsthat AKB = 90 + . Inside equilateral triangle AOB there exists a unique point K thatserves as the vertex of the angles that subtend its sides and are equal to the given angles.

Similar arguments for a point L inside triangle COD show that OKB = CLO.Now, let us prove that KOL = OKB. Indeed, COL = KBO; hence, KOB +

COL = 180 OKB = 90 and, therefore, KOL = 2 + (90 ) = 90 + =OKB. It follows that KL = OB = R. Similarly, LM =MK = R.

1.45. Let A = . It is easy to verify that both angles KCL and ADL are equal to240 (or 120 + ). Since KC = BC = AD and CL = DL, it follows that KCL =ADL and, therefore, KL = AL. Similarly, KL = AK.

1.46. Let P , Q and R be the centers of the squares constructed on sides DA, AB andBC, respectively, in parallelogram ABCD with an acute angle of at vertex A. It is easyto verify that PAQ = 90 + = RBQ; hence, PAQ = RBQ. Sides AQ and BQ ofthese triangles are perpendicular, hence, PQ QR.

1.47. First, observe that the sum of the angles at vertices A, B and C of hexagonABCABC is equal to 360 because by the hypothesis the sum of its angles at the othervertices is equal to 360. On side AC , construct outwards triangle AC P equal to triangleBC A (Fig. 10).


Then ABP = CBA because AB = CB, AP = CA andPAB = 360 PAC C AB = 360 ABC C AB = ACB.

Hence, C BA = C BP and, therefore, 2ABC = PBA = ABC becausePBA = ABC.

SOLUTIONS 29

1.48. Since BA : BC = BC1 : BA1 and ABC = C1BA1, it follows that ABC C1BA1. Similarly, ABC B1A1C. Since BA1 = A1C, it follows that C1BA1 =B1A1C. Therefore, AC1 = C1B = B1A1 and AB1 = B1C = C1A1. It is also clear thatquadrilateral AB1A1C1 is a convex one.

1.49. a) Let P and Q be the midpoints of sides AB and AC. Then MP = 12AC = QB1,

MQ = 12AB = PC1 and C1PM = C1PB + BPM = B1QC + CQM = B1QM .

Hence, MQB1 = C1PM and, therefore, MC1 =MB1. Moreover,PMC1 + QMB1 = QB1M + QMB1 = 180

MQB1and

MQB1 = A+ CQB1 = A+ (180 2).

Therefore, B1MC1 = PMQ+2A = 2. (The case when C1PB+BPM > 180is analogously treated.)

b) On sides AB and AC, take points B and C , respectively, such that AB : AB = AC :AC = 2 : 3. The midpoint M of segment BC coincides with the intersection point of themedians of triangle ABC. On sides AB and AC , construct outwards right triangles ABC1and AB1C

with angle = 60 as in heading a). Then B1 and C1 are the centers of righttriangles constructed on sides AB and AC; on the other hand, by heading a), MB1 =MC1and B1MC1 = 120

.

Remark. Statements of headings a) and b) remain true for triangles constructed in-wards, as well.

1.50. a) Let B be the intersection point of line AC and the perpendicular to line AB1erected from point B1; define point C

similarly. Since AB : AC = AC1 : AB1 = AB : AC,it follows that BC BC. If N is the midpoint of segment BC , then, as follows fromProblem 1.49, NC1 = NB1 (i.e., N =M) and B1NC1 = 2AB

B1 = 1802CAB1 = .b) On side BC construct outwards isosceles triangle BA1C with angle 360

2 at vertexA1 (if < 90

construct inwards a triangle with angle 2). Since the sum of the angles atthe vertices of the three constructed isosceles triangles is equal to 360, it follows that theangles of triangle A1B1C1 are equal to 180

, 12 and 1

2 (cf. Problem 1.47). In particular,

this triangle is an isosceles one, hence, A1 = O.1.51. Let O1, O2, O3 and O4 be the centers of rhombuses constructed on sides AB, BC,

CA and DA, respectively; let M be the midpoint of diagonal AC. Then MO1 = MO2 andO1MO2 = (cf. Problem 1.49). Similarly, MO3 = MO4 and O3MO4 = . Therefore,under the rotation through an angle of about point M triangle O1MO3 turns intoO2MO4.

1.52. Since A1C = AC| cosC| , B1C = BC| cosC| and angle C is the common angleof triangles ABC and A1B1C, these triangles are similar; the similarity coefficient is equalto | cosC|.

1.53. Since pointsM and N lie on the circle with diameter CH, it follows that CMN =CHN and since AC HN , we see that CHN = A. Similarly, CNM = B.

1.54. a) Let l be the tangent to the circumscribed circle at point A. Then (l, AB) =(AC,CB) = (C1B1, AC1) and, therefore, l B1C1.

b) Since OA l and l B1C1, it follows that OA B1C1.1.55. If AA1, BB1 and CC1 are heights, then right triangles AA1C and BB1C have

equal angles at vertex C and, therefore, are similar. It follows that A1BH B1AH,consequently, AH A1H = BH B1H. Similarly, BH B1H = CH C1H.

If AH A1H = BH B1H = CH C1H, then A1BH B1AH; hence, BA1H =AB1H = . Thus, CA1H = CB1H = 180

.


Similarly, AC1H = CA1H = 180 and AC1H = AB1H = . Hence, = 90,

i.e., AA1, BB1 and CC1 are heights.1.56. a) By Problem 1.52 C1A1B = CA1B1 = A. Since AA1 BC, it follows that

C1A1A = B1A1A. The proof of the fact that rays B1B and C1C are the bisectors ofangles A1B1C1 and A1C1B1 is similar.

b) Lines AB, BC and CA are the bisectors of the outer angles of triangle A1B1C1, hence,A1A is the bisector of angle B1A1C1 and, therefore, AA1 BC. For lines BB1 and CC1the proof is similar.

1.57. From the result of Problem 1.56 a) it follows that the symmetry through line ACsends line B1A1 into line B1C1.

1.58. By Problem 1.52 B1A1C = BAC. Since A1B1 AB, it follows that B1A1C =ABC. Hence, BAC = ABC. Similarly, since B1C1 BC, it follows that ABC =BCA. Therefore, triangle ABC is an equilateral one and A1C1 AC.

1.59. Let O be the center of the circumscribed circle of triangle ABC. Since OA B1C1(cf. Problem 1.54 b), it follows that SAOC1 + SAOB1 =

12(R B1C1). Similar arguments for

vertices B and C show that SABC = qR. On the other hand, SABC = pr.1.60. The perimeter of the triangle cut off by the line parallel to side BC is equal to

the sum of distances from point A to the tangent points of the inscribed circle with sidesAB and AC; therefore, the sum of perimeters of small triangles is equal to the perimeterof triangle ABC, i.e., P1 + P2 + P3 = P . The similarity of triangles implies that

rir= Pi

P.

Summing these equalities for all the i we get the statement desired.1.61. Let M = A. Then XA = A; hence, AYA = 1. Similarly, CXC = 1. Let us

prove that y = AYA and x = CXC are the desired lines. On side BC, take point D so thatAB MD, see Fig. 11. Let E be the intersection point of lines CXC and MD. Then,XMM + YMM = XCE + YMM . Since ABC MDC, it follows that CE = YMM .Therefore, CE = YMM . Hence, XMM + YMM = XCE + CE = XCC = 1.


1.62. Let D be the midpoint of segment BH. Since BHA HEA, it follows thatAD : AO = AB : AH and DAH = OAE. Hence, DAO = BAH and, therefore,DAO BAH and DOA = BAH = 90.

1.63. Let AA1, BB1 and CC1 be heights of triangle ABC. Let us drop from point B1perpendicularsB1K and B1N to sides AB and BC, respectively, and perpendicularsB1L andB1M to heights AA1 and CC1, respectively. Since KB1 : C1C = AB1 : AC = LB1 : A1C,it follows that KLB1 C1A1C and, therefore, KL C1A1. Similarly, MN C1A1.Moreover, KN C1A1 (cf. Problem 1.53). It follows that points K, L, M and N lie on oneline.

1.64. a) Let O be the midpoint of AC, let O1 be the midpoint of AB and O2 the midpointof BC. Assume that AB BC. Through point O1 draw line O1K parallel to EF (point Klies on segment EO2). Let us prove that right triangles DBO and O1KO2 are equal. Indeed,

SOLUTIONS 31

O1O2 = DO =12AC and BO = KO2 =

12(BC AB). Since triangles DBO and O1KO2

are equal, we see that BOD = O1O2E, i.e., line DO is parallel to EO2 and the tangentdrawn through point D is parallel to line EF .

b) Since the angles between the diameter AC and the tangents to the circles at points F ,D, E are equal, it follows that FAB = DAC = EBC and FBA+DCA = ECB,i.e., F lies on line segment AD and E lies on line segment DC. Moreover, AFB =BEC = ADC = 90 and, therefore, FDEB is a rectangle.

1.65. Let MQ and MP be perpendiculars dropped on sides AD and BC, let MR andMT be perpendiculars dropped on the extensions of sides AB and CD (Fig. 12). Denoteby M1 and P1 the other intersection points of lines RT and QP with the circle.


Since TM1 = RM = AQ and TM1 AQ, it follows that AM1 TQ. Similarly, AP1 RP . Since M1AP1 = 90

, it follows that RP TQ.Denote the intersection points of lines TQ and RP , M1A and RP , P1A and TQ by E,

F , G, respectively. To prove that point E lies on line AC, it suffices to prove that rectanglesAFEG and AM1CP1 are similar. Since ARF = AM1R = M1TG = M1CT , wemay denote the values of these angles by the same letter . We have: AF = RA sin =M1A sin

2 and AG = M1T sin = M1C sin2 . Therefore, rectangles AFEG and AM1CP1

are similar.1.66. Denote the centers of the circles by O1 and O2. The outer tangent is tangent to

the first circle at point K and to the other circle at point L; the inner tangent is tangent tothe first circle at point M and to the other circle at point N (Fig. 13).


Let lines KM and LN intersect line O1O2 at points P1 and P2, respectively. We haveto prove that P1 = P2. Let us consider points A, D1, D2 the intersection points of KLwithMN , KM with O1A, and LN with O2A, respectively. Since O1AM +NAO2 = 90

,


right triangles O1MA and ANO2 are similar; we also see that AO2 KM and AO1 LN .Since these lines are parallel, AD1 : D1O1 = O2P1 : P1O1 and D2O2 : AD2 = O2P2 : P2O1.The similarity of quadrilaterals AKO1M and O2NAL yields AD1 : D1O1 = D2O2 : AD2.Therefore, O2P1 : P1O1 = O2P2 : P2O1, i.e., P1 = P2.

CHAPTER 2. INSCRIBED ANGLES

Background

1. Angle ABC whose vertex lies on a circle and legs intersect this circle is calledinscribed in the circle. Let O be the center of the circle. Then

ABC =

{12AOC if points B and O lie on one side of AC

180 12AOC otherwise.

The most important and most often used corollary of this fact is that equal chords subtendangles that either are equal or the sum of the angles is equal to 180.

2. The value of the angle between chord AB and the tangent to the circle that passesthrough point A is equal to half the angle value of arc AB.

3. The angle values of arcs confined between parallel chords are equal.4. As we have already said, if two angles subtend the same chord, either they are equal

or the sum of their values is 180. In order not to consider various variants of the positionsof points on the circle let us introduce the notion of an oriented angle between lines. Thevalue of the oriented angle between lines AB and CD (notation: (AB,CD)) is the valueof the angle by which we have to rotate line AB counterclockwise in order for it to becomeparallel to line CD. The angles that differ by n 180 are considered equal.

Notice that, generally, the oriented angle between lines CD and AB is not equal to theoriented angle between lines AB and CD (the sum of (AB,CD) and (CD,AB) is equalto 180 which, according to our convention, is the same as 0).

It is easy to verify the following properties of the oriented angles:a) (AB,BC) = (BC,AB);b) (AB,CD) + (CD,EF ) = (AB,EF );c) points A, B, C, D not on one line lie on one circle if and only if (AB,BC) =

(AD,DC). (To prove this property we have to consider two cases: points B and D lie onone side of AC; points B and D lie on different sides of AC.)

Introductory problems

1. a) From point A lying outside a circle rays AB and AC come out and intersect thecircle. Prove that the value of angle BAC is equal to half the difference of the anglemeasures of the arcs of the circle confined inside this angle.

b) The vertex of angle BAC lies inside a circle. Prove that the value of angle BAC isequal to half the sum of angle measures of the arcs of the circle confined inside angle BACand inside the angle symmetric to it through vertex A.

2. From point P inside acute angle BAC perpendiculars PC1 and PB1 are droppedon lines AB and AC. Prove that C1AP = C1B1P .

3. Prove that all the angles formed by the sides and diagonals of a regular n-gon areinteger multiples of 180

n.

4. The center of an inscribed circle of triangle ABC is symmetric through side AB tothe center of the circumscribed circle. Find the angles of triangle ABC.

33

34 CHAPTER 2. INSCRIBED ANGLES

5. The bisector of the exterior angle at vertex C of triangle ABC intersects the circum-scribed circle at point D. Prove that AD = BD.

1. Angles that subtend equal arcs

2.1. Vertex A of an acute triangle ABC is connected by a segment with the center O ofthe circumscribed circle. From vertex A height AH is drawn. Prove that BAH = OAC.

2.2. Two circles intersect at points M and K. Lines AB and CD are drawn through Mand K, respectively; they intersect the first circle at points A and C, the second circle atpoints B and D, respectively. Prove that AC BD.

2.3. From an arbitrary point M inside a given angle with vertex A perpendiculars MPand MQ are dropped to the sides of the angle. From point A perpendicular AK is droppedon segment PQ. Prove that PAK = MAQ.

2.4. a) The continuation of the bisector of angle B of triangle ABC intersects thecircumscribed circle at point M ; O is the center of the inscribed circle, Ob is the center ofthe escribed circle tangent to AC. Prove that points A, C, O and Ob lie on a circle centeredat M .

b) Point O inside triangle ABC is such that lines AO, BO and CO pass through thecenters of the circumscribed circles of triangles BCO, ACO and ABO, respectively. Provethat O is the center of the inscribed circle of triangle ABC.

2.5. Vertices A and B of right triangle ABC with right angle C slide along the sidesof a right angle with vertex P . Prove that in doing so point C moves along a line segment.

2.6. Diagonal AC of square ABCD coincides with the hypothenuse of right triangleACK, so that points B and K lie on one side of line AC. Prove that

BK =|AK CK|

2and DK =

AK + CK2

.

2.7. In triangle ABC medians AA1 and BB1 are drawn. Prove that if CAA1 =CBB1, then AC = BC.

2.8. Each angle of triangle ABC is smaller than 120. Prove that inside ABC thereexists a point that serves as the vertex for three angles each of value 120 and subtendingthe side of the triangle different from the sides subtended by the other angles.

2.9. A circle is divided into equal arcs by n diameters. Prove that the bases of theperpendiculars dropped from an arbitrary point M inside the circle to these diameters arevertices of a regular n-gon.

2.10. Points A, B, M and N on a circle are given. From pointM chordsMA1 andMB1perpendicular to lines NB and NA, respectively, are drawn. Prove that AA1 BB1.

2.11. Polygon ABCDEF is an inscribed one; AB DE and BC EF . Prove thatCD AF .

2.12. Polygon A1A2 . . . A2n as an inscribed one. We know that all the pairs of its oppositesides except one are parallel. Prove that for any odd n the remaining pair of sides is alsoparallel and for any even n the lengths of the exceptional sides are equal.

2.13. Consider triangle ABC. Prove that there exist two families of equilateral triangleswhose sides (or extensions of the sides) pass through points A, B and C. Prove also thatthe centers of triangles from these families lie on two concentric circles.

3. THE ANGLE BETWEEN A TANGENT AND A CHORD 35

2. The value of an angle between two chords

The following fact helps to solve problems from this section. Let A, B, C, D be pointson a circle situated in the order indicated. Then

(AC,BD) = AB+ CD

2and (AB,CD) =

| AD CB|2

.

To prove this, we have to draw a chord parallel to another chord through the endpoint ofone of the chords.

2.14. Points A, B, C, D in the indicated order are given on a circle. Let M be themidpoint of arc AB. Denote the intersection points of chords MC and MD with chordAB by E and K. Prove that KECD is an inscribed quadrilateral.

2.15. Concider an equilateral triangle. A circle with the radius equal to the trianglesheight rolls along a side of the triangle. Prove that the angle measure of the arc cut off thecircle by the sides of the triangle is always equal to 60.

2.16. The diagonals of an isosceles trapezoid ABCD with lateral side AB intersect atpoint P . Prove that the center O of the inscribed circle lies on the inscribed circle of triangleAPB.

2.17. Points A, B, C, D in the indicated order are given on a circle; points A1, B1, C1and D1 are the midpoints of arcs AB, BC, CD and DA, respectively. Provethat A1C1 B1D1.

2.18. Point P inside triangle ABC is taken so that BPC = A + 60, APC =B + 60 and APB = C + 60. Lines AP , BP and CP intersect the circumscribedcircle of triangle ABC at points A, B and C , respectively. Prove that triangle ABC isan equilateral one.

2.19. Points A, C1, B, A1, C, B1 in the indicated order are taken on a circle.a) Prove that if lines AA1, BB1 and CC1 are the bisectors of the angles of triangle ABC,

then they are the heights of triangle A1B1C1.b) Prove that if lines AA1, BB1 and CC1 are the heights of triangle ABC, then they are

the bisectors of the angles of triangle A1B1C1.2.20. Triangles T1 and T2 are inscribed in a circle so that the vertices of triangle T2

are the midpoints of the arcs into which the circle is divided by the vertices of triangle T1.Prove that in the hexagon which is the intersection of triangles T1 and T2 the diagonals thatconnect the opposite vertices are parallel to the sides of triangle T1 and meet at one point.

3. The angle between a tangent and a chord

2.21. Two circles intersect in points P and Q. Through point A on the first circle linesAP and AQ are drawn. The lines intersect the second circle in points B and C. Prove thatthe tangent at A to the first circle is parallel to line BC.

2.22. Circles S1 and S2 intersect at points A and P . Tangent AB to circle S1 is drawnthrough point A, and line CD parallel to AB is drawn through point P (points B and C lieon S2, point D on S1). Prove that ABCD is a parallelogram.

2.23. The tangent at point A to the inscribed circle of triangle ABC intersects line BCat point E; let AD be the bisector of triangle ABC. Prove that AE = ED.

2.24. Circles S1 and S2 intersect at point A. Through point A a line that intersects S1at point B and S2 at point C is drawn. Through points C and B tangents to the circles aredrawn; the tangents intersect at point D. Prove that angle BDC does not depend on thechoice of the line that passes through A.

2.25. Two circles intersect at points A and B. Through point A tangents AM and AN ,where M and N are points of the respective circles, are drawn. Prove that:


a) ABN + MAN = 180;b) BM

BN=(AMAN

)2.

2.26. Inside square ABCD a point P is taken so that triangle ABP is an equilateralone. Prove that PCD = 15.

2.27. Two circles are internally tangent at point M . Let AB be the chord of the greatercircle which is tangent to the smaller circle at point T . Prove that MT is the bisector ofangle AMB.

2.28. Through pointM inside circle S chord AB is drawn; perpendicularsMP andMQare dropped from point M to the tangents that pass through points A and B respectively.Prove that the value of 1

PM+ 1

QMdoes not depend on the choice of the chord that passes

through point M .2.29. Circle S1 is tangent to sides of angle ABC at points A and C. Circle S2 is tangent

to line AC at point C and passes through point B, circle S2 intersects circle S1 at point M .Prove that line AM divides segment BC in halves.

2.30. Circle S is tangent to circles S1 and S2 at points A1 and A2; let B be a point ofcircle S, let K1 and K2 be the other intersection points of lines A1B and A2B with circles S1and S2, respectively. Prove that if line K1K2 is tangent to circle S1, then it is also tangentto circle S2.

4. Relations between the values of an angle and the lengths of the arc andchord associated with the angle

2.31. Isosceles trapezoids ABCD and A1B1C1D1 with parallel respective sides are in-scribed in a circle. Prove that AC = A1C1.

2.32. From pointM that moves along a circle perpendicularsMP andMQ are droppedon diameters AB and CD, respectively. Prove that the length of segment PQ does notdepend on the position of point M .

2.33. In triangle ABC, angle B is equal to 60; bisectors AD and CE intersect atpoint O. Prove that OD = OE.

2.34. In triangle ABC the angles at vertices B and C are equal to 40; let BD be thebisector of angle B. Prove that BD +DA = BC.

2.35. On chord AB of circle S centered at O a point C is taken. The circumscribedcircle of triangle AOC intersects circle S at point D. Prove that BC = CD.

2.36. Vertices A and B of an equilateral triangle ABC lie on circle S, vertex C liesinside this circle. Point D lies on circle S and BD = AB. Line CD intersects S at point E.Prove that the length of segment EC is equal to the radius of circle S.

2.37. Along a fixed circle another circle whose radius is half that of the fixed one rollson the inside without gliding. What is the trajectory of a fixed point K of the rolling circle?

5. Four points on one circle

2.38. From an arbitrary point M on leg BC of right triangle ABC perpendicular MNis dropped on hypothenuse AP . Prove that MAN = MCN .

2.39. The diagonals of trapezoid ABCD with bases AD and BC intersect at point O;points B and C are symmetric through the bisector of angle BOC to vertices B and C,respectively. Prove that C AC = BDB.

2.40. The extensions of sides AB and CD of the inscribed quadrilateral ABCD meet atpoint P ; the extensions of sides BC and AD meet at point Q. Prove that the intersectionpoints of the bisectors of angles AQB and BPC with the sides of the quadrilateral arevertices of a rhombus.

6. THE INSCRIBED ANGLE AND SIMILAR TRIANGLES 37

2.41. The inscribed circle of triangle ABC is tangent to sides AB and AC at points Mand N , respectively. Let P be the intersection point of line MN with the bisector (or itsextension) of angle B. Prove that:

a) BPC = 90;b) SABP : SABC = 1 : 2.2.42. Inside quadrilateral ABCD a pointM is taken so that ABMD is a parallelogram.

Prove that if CBM = CDM , then ACD = BCM .2.43. Lines AP , BP and CP intersect the circumscribed circle of triangle ABC at

points A1, B1 and C1, respectively. On lines BC, CA and AB points A2, B2 and C2,respectively, are taken so that (PA2, BC) = (PB2, CA) = (PC2, AB). Prove thatA2B2C2 A1B1C1.

2.44. About an equilateral triangle APQ a rectangular ABCD is circumscribed so thatpoints P and Q lie on sides BC and CD, respectively; P and Q are the midpoints of sidesAP and AQ, respectively. Prove that triangles BQC and CP D are equilateral ones.

2.45. Prove that if for inscribed quadrilateral ABCD the equality CD = AD + BCholds, then the intersection point of the bisectors of angles A and B lies on side CD.

2.46. Diagonals AC and CE of a regular hexagon ABCDEF are divided by points Mand N , respectively, so that AM : AC = CN : CE = . Find if it is known that pointsB, M and N lie on a line.

2.47. The corresponding sides of triangles ABC and A1B1C1 are parallel and sides ABand A1B1 lie on one line. Prove that the line that connects the intersection points of thecircumscribed circles of triangles A1BC and AB1C contains point C1.

2.48. In triangle ABC heights AA1, BB1 and CC1 are drawn. Line KL is parallel toCC1; points K and L lie on lines BC and B1C1, respectively. Prove that the center of thecircumscribed circle of triangle A1KL lies on line AC.

2.49. Through the intersection point O of the bisectors of triangle ABC line MN isdrawn perpendicularly to CO so that M and N lie on sides AC and BC, respectively.Lines AO and BO intersect the circumscribed circle of triangle ABC at points A and B,respectively. Prove that the intersection point of lines AN andBM lies on the circumscribedcircle.

6. The inscribed angle and similar triangles

2.50. Points A, B, C and D on a circle are given. Lines AB and CD intersect at pointM . Prove that

AC ADAM

=BC BDBM

.

2.51. Points A, B and C on a circle are given; the distance BC is greater than thedistance from point B to line l tangent to the circle at point A. Line AC intersects the linedrawn through point B parallelly to l at point D. Prove that AB2 = AC AD.

2.52. Line l is tangent to the circle of diameter AB at point C; points M and N are theprojections of points A and B on line l, respectively, and D is the projection of point C onAB. Prove that CD2 = AM BN .

2.53. In triangle ABC, height AH is drawn and from vertices B and C perpendicularsBB1 and CC1 are dropped on the line that passes through point A. Prove that ABC HB1C1.

2.54. On arc BC of the circle circumscribed about equilateral triangle ABC, pointP is taken. Segments AP and BC intersect at point Q. Prove that

1

PQ=

1

PB+

1

PC.


2.55. On sides BC and CD of square ABCD points E and F are taken so that EAF =45. Segments AE and AF intersect diagonal BD at points P and Q, respectively. Provethat SAEF

SAPQ= 2.

2.56. A line that passes through vertex C of equilateral triangle ABC intersects baseAB at point M and the circumscribed circle at point N . Prove that

CM CN = AC2 and CMCN

=AM BMAN BN .

2.57. Consider parallelogram ABCD with an acute angle at vertex A. On rays AB andCB points H and K, respectively, are marked so that CH = BC and AK = AB. Provethat:

a) DH = DK;b) DKH ABK.2.58. a) The legs of an angle with vertex C are tangent to a circle at points A and B.

From point P on the circle perpendiculars PA1, PB1 and PC1 are dropped on lines BC,CA and AB, respectively. Prove that PC21 = PA1 PB1.

b) From point O of the inscribed circle of triangle ABC perpendiculars OA, OB, OC

are dropped on the sides of triangle ABC opposite to vertices A, B and C, respectively, andperpendiculars OA, OB, OC are dropped to the sides of the triangle with vertices at thetangent points. Prove that

OA OB OC = OA OB OC .2.59. Pentagon ABCDE is inscribed in a circle. Distances from point E to lines AB,

BC and CD are equal to a, b and c, respectively. Find the distance from point E to lineAD.

2.60. In triangle ABC, heights AA1, BB1 and CC1 are drawn; B2 and C2 are themidpoints of heights BB1 and CC1, respectively. Prove that A1B2C2 ABC.

2.61. On heights of triangle ABC points A1, B1 and C1 that divide them in the ratio2 : 1 counting from the vertex are taken. Prove that A1B1C1 ABC.

2.62. Circle S1 with diameter AB intersects circle S2 centered at A at points C andD. Through point B a line is drawn; it intersects S2 at point M that lies inside S1 and itintersects S1 at point N . Prove that MN

2 = CN ND.2.63. Through the midpoint C of an arbitrary chord AB on a circle chords KL andMN

are drawn so that points K and M lie on one side of AB. Segments KN and ML intersectAB at points Q and P , respectively. Prove that PC = QC.

2.64. a) A circle that passes through point C intersects sides BC and AC of triangleABC at points A1 and B1, respectively, and it intersects the circumscribed circle of triangleABC at point M . Prove that AB1M BA1M .

b) On rays AC and BC segments AA1 and BB1 equal to the semiperimeter of triangleABC are drawn. LetM be a point on the circumscribed circle such that CM A1B1. Provethat CMO = 90, where O is the center of the inscribed circle.

7. The bisector divides an arc in halves

2.65. In triangle ABC, sides AC and BC are not equal. Prove that the bisector of angleC divides the angle between the median and the height drawn from this vertex in halvesif and only if C = 90.

2.66. It is known that in a triangle the median, the bisector and the height drawn fromvertex C divide the angle C into four equal parts. Find the angles of this triangle.

2.67. Prove that in triangle ABC bisector AE lies between median AM and height AH.

9. THREE CIRCUMSCRIBED CIRCLES INTERSECT AT ONE POINT 39

2.68. Given triangle ABC; on its side AB point P is chosen; lines PM and PN parallelto AC and BC, respectively, are drawn through P so that points M and N lie on sides BCand AC, respectively; let Q be the intersection point of the circumscribed circles of trianglesAPN and BPM . Prove that all lines PQ pass through a fixed point.

2.69. The continuation of bisector AD of acute triangle ABC inersects the circumscribedcircle at point E. Perpendiculars DP and DQ are dropped on sides AB and AC from pointD. Prove that SABC = SAPEQ.

8. An inscribed quadrilateral with perpendicular diagonals

In this section ABCD is an inscribed quadrilateral whose diagonals intersect at a rightangle. We will also adopt the following notations: O is the center of the circumscribed circleof quadrilateral ABCD and P is the intersection point of its diagonals.

2.70. Prove that the broken line AOC divides ABCD into two parts whose areas areequal.

2.71. The radius of the circumscribed circle of quadrilateral ABCD is equal to R.a) Find AP 2 +BP 2 + CP 2 +DP 2.b) Find the sum of squared lengths of the sides of ABCD.2.72. Find the sum of squared lengths of the diagonals of ABCD if the length of segment

OP and the radius of the circumscribed circle R are known.2.73. From vertices A and B perpendiculars to CD that intersect lines BD and AC at

points K and L, respectively, are drawn. Prove that AKLB is a rhombus.2.74. Prove that the area of quadrilateral ABCD is equal to 1

2(AB CD +BC AD).

2.75. Prove that the distance from point O to side AB is equal to half the length of sideCD.

2.76. Prove that the line drawn through point P perpendicularly to BC divides sideAD in halves.

2.77. Prove that the midpoints of the sides of quadrilateral ABCD and the projectionsof point P on the sides lie on one circle.

2.78. a) Through vertices A, B, C and D tangents to the circumscribed circle are drawn.Prove that the quadrilateral formed by them is an inscribed one.

b) Quadrilateral KLMN is simultaneously inscribed and circumscribed; A and B arethe tangent points of the inscribed circle with sides KL and LM , respectively. Prove thatAK BM = r2, where r is the radius of the inscribed circle.

9. Three circumscribed circles intersect at one point

2.79. On sides of triangle ABC triangles ABC , ABC and ABC are constructedoutwards so that the sum of the angles at vertices A, B and C is a multiple of 180. Provethat the circumscribed circles of the constructed triangles intersect at one point.

2.80. a) On sides (or their extensions) BC, CA and AB of triangle ABC points A1, B1and C1 distinct from the vertices of the triangle are taken (one point on one side). Provethat the circumscribed circles of triangles AB1C1, A1BC1 and A1B1C intersect at one point.

b) Points A1, B1 and C1 move along lines BC, CA and AB, respectively, so that alltriangles A1B1C1 are similar and equally oriented. Prove that the intersection point of thecircumscribed circles of triangles AB1C1, A1BC1 and A1B1C remains fixed in the process.

2.81. On sides BC, CA and AB of triangle ABC points A1, B1 and C1 are taken.Prove that if triangles A1B1C1 and ABC are similar and have opposite orientations, thencircumscribed circles of triangles AB1C1, ABC1 and A1B1C pass through the center of thecircumscribed circle of triangle ABC.


2.82. Points A, B and C are symmetric to a point P relative sides BC, CA and AB,respectively, of triangle ABC.

a) The circumscribed circles of triangles ABC , ABC , ABC and ABC have a commonpoint;

b) the circumscribed circles of triangles ABC, ABC, ABC and ABC have a commonpoint Q;

c) Let I, J , K and O be the centers of the circumscribed circles of triangles ABC, ABC,ABC and ABC , respectively. Prove that QI : OI = QJ : OJ = QK : OK.

10. Michels point

2.83. Four lines form four triangles. Prove thata) The circumscribed circles of these triangles have a common point. (Michels point.)b) The centers of the circumscribed circles of these triangles lie on one circle that passes

through Michels point.2.84. A line intersects sides (or their extensions) AB, BC and CA of triangle ABC at

points C1, B1 and A1, respectively; let O, Oa, Ob and Oc be the centers of the circumscribedcircles of triangles ABC, AB1C1, A1BC1 and A1B1C, respectively; let H, Ha, Hb and Hc bethe respective orthocenters of these triangles. Prove that

a) OaObOc ABC.b) the midperpendiculars to segments OH, OaHa, ObHb and OcHc meet at one point.2.85. Quadrilateral ABCD is an inscribed one. Prove that Michels point of lines that

contain its sides lies on the segment that connects the intersection points of the extensionsof the sides.

2.86. Points A, B, C and D lie on a circle centered at O. Lines AB and CD intersectat point E and the circumscribed circles of triangles AEC and BED ilntersect at points Eand P . Prove that

a) points A, D, P and O lie on one circle;b) EPO = 90.2.87. Given four lines prove that the projections of Michels point to these lines lie on

one line.

See also Problem 19.45.

11. Miscellaneous problems

2.88. In triangle ABC height AH is drawn; let O be the center of the circumscribedcircle. Prove that OAH = |B C|.

2.89. Let H be the intersection point of the heights of triangle ABC; let AA be adiameter of its circumscribed circle. Prove that segment AH divides side BC in halves.

2.90. Through vertices A and B of triangle ABC two parallel lines are drawn and linesm and n are symmetric to them through the bisectors of the corresponding angles. Provethat the intersection point of lines m and n lies on the circumscribed circle of triangle ABC.

2.91. a) Lines tangent to circle S at points B

planegeo.pdf

Documents

introductory problems

miscellaneous problems

known problems

new problems

areas of similar triangles

triangles heights

auxiliary equal triangles

regrouping areas